Photoelectric effect calculation

[*now*]

Photoelectric effect
1. Homework Statement
A metal surface is illuminated with monochromatic light and it becomes charged to a steady positive potential of 1.0V relative to its surroundings. The work function energy of the metal surface is 3.0eV, the electron charge is 1.6 x 10-19 C. Calculate the frequency of the light.

Homework Equations

hf=work function + max kinetic energy of an electron. E=hf

The Attempt at a Solution

wavespeed/wavelength = frequency. work function =3.0 eV electron charge = 1.6x10-19

 

[gneill]

Hi "now",

You haven't made an attempt at solution. What are your thoughts about how to approach the problem? What are the implications of "charged to a steady positive potential" if the metal surface continues to be illuminated? That is, what does it mean for electrons that absorb the light photons?

 

[*now*]

Hi, I'm sorry, I'm having a lot of trouble understanding what is asked. We had our first lesson about the photoelectric effect today. Does 'charged to a steady positive potential' mean it isn't changing?

 

[gneill]

Hi, I'm sorry, I'm having a lot of trouble understanding what is asked. We had our first lesson about the photoelectric effect today. Does 'charged to a steady positive potential' mean it isn't changing?

Yes. It rises to some value as electrons are "kicked" off of the surface by the photoelectric effect. Why do you think it reaches a steady value?

 

[*now*]

Yes. It rises to some value as electrons are "kicked" off of the surface by the photoelectric effect. Why do you think it reaches a steady value?

It's monochromatic light so the value stays the same, or does it increase? Do you need to consider the speed of light?

 

[gneill]

It's monochromatic light so the value stays the same, or does it increase? Do you need to consider the speed of light?

I'm not sure what you're thinking here. Perhaps you should try describing in your own words what happens, starting with the plate in an uncharged state and the light source is first directed at the metal surface. While you do this, think about the potentials and energies involved in the photoelectric effect.

 

[*now*]

I think I needed to use the equation frequency is wave speed divided by wavelength thus 3.8x10^8 over 4.0 x10-7^-7 equalling 7.5 x10^14 Hz. This times planks constant is 4.95x10^-19 which is the answer for the amount of energy. My work has been marked but I want to check my thinking with you, does this seem right?

 

[gneill]

I think I needed to use the equation frequency is wave speed divided by wavelength thus 3.8x10^8 over 4.0 x10-7^-7 equalling 7.5 x10^14 Hz. This times planks constant is 4.95x10^-19 which is the answer for the amount of energy. My work has been marked but I want to check my thinking with you, does this seem right?

You probably don't need the f = c/λ formula with the values that were given to you in the problem. But even so, where does your value of 4.0 x 10-7 come from? I don't see it as a given value in the problem statement.

I would look at the energy required for an electron to escape the plate when the plate's potential reaches its equilibrium state (1.0 V with respect to its surroundings).

 

[*now*]

Sorry wrote wrong answer, had to get the energy by multiplying q and v to get 1.6x10^-19. Then I used the equation hf= work function + KEmax which goes to 4.8 x10^-19 (3eV) + 1.6 x 10^-19 which gives 6.4 x 10^-19. To find frequency we divide this by planks constant which gives 9.69 x 10^14Hz.